Improper integrals with polynomial in the denominator - how does p-series apply?
up vote
4
down vote
favorite
For example,
$$int_2^{infty} frac1{x^2+2x-3} dx$$
The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?
calculus integration improper-integrals
asked Nov 27 at 0:38
smxx
211
add a comment |
up vote
4
down vote
favorite
For example,
$$int_2^{infty} frac1{x^2+2x-3} dx$$
The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?
calculus integration improper-integrals
asked Nov 27 at 0:38
smxx
211
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
For example,
$$int_2^{infty} frac1{x^2+2x-3} dx$$
The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?
calculus integration improper-integrals
asked Nov 27 at 0:38
smxx
211
For example,
$$int_2^{infty} frac1{x^2+2x-3} dx$$
The power in the denominator is greater than 1, so shouldn't it converge? When solving with partial fractions, it appears that it diverges. However, WolframAlpha gives me a numerical answer (log5/4). Do I need to solve it with a different method? Does it converge of diverge?
calculus integration improper-integrals
calculus integration improper-integrals
asked Nov 27 at 0:38
smxx
211
asked Nov 27 at 0:38
smxx
211
asked Nov 27 at 0:38
smxx
211
asked Nov 27 at 0:38
smxx
211
asked Nov 27 at 0:38
smxx
211
211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
begin{align}
int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
& = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
& = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
& = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
& = -frac 14 ln Big(frac 15Big) \
& = frac 14 ln5
end{align}
Basically, there are many subtle things going on here.
For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".
This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$
We know that
$$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$
Hence
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$
which is where the "infinities cancel out".
This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that
$$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$
However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.
answered Nov 27 at 0:50
glowstonetrees
2,195317
add a comment |
up vote
0
down vote
When you use partial fractions you may loose convergence on each single fraction.
To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.
answered Nov 27 at 0:55
G Cab
17.4k31237
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015158%2fimproper-integrals-with-polynomial-in-the-denominator-how-does-p-series-apply%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
begin{align}
int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
& = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
& = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
& = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
& = -frac 14 ln Big(frac 15Big) \
& = frac 14 ln5
end{align}
Basically, there are many subtle things going on here.
For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".
This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$
We know that
$$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$
Hence
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$
which is where the "infinities cancel out".
This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that
$$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$
However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.
answered Nov 27 at 0:50
glowstonetrees
2,195317
add a comment |
up vote
3
down vote
begin{align}
int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
& = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
& = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
& = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
& = -frac 14 ln Big(frac 15Big) \
& = frac 14 ln5
end{align}
Basically, there are many subtle things going on here.
For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".
This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$
We know that
$$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$
Hence
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$
which is where the "infinities cancel out".
This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that
$$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$
However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.
answered Nov 27 at 0:50
glowstonetrees
2,195317
add a comment |
up vote
3
down vote
up vote
3
down vote
begin{align}
int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
& = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
& = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
& = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
& = -frac 14 ln Big(frac 15Big) \
& = frac 14 ln5
end{align}
Basically, there are many subtle things going on here.
For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".
This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$
We know that
$$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$
Hence
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$
which is where the "infinities cancel out".
This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that
$$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$
However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.
answered Nov 27 at 0:50
glowstonetrees
2,195317
begin{align}
int_2^infty frac{1}{x^2+2x-3} dx & = frac 14 int_2^infty bigg(frac{1}{x-1} - frac{1}{x+3} bigg) dx \ tag{1}
& = frac 14 bigg[ln|x-1| - ln |x+3| bigg]_2^infty \ tag{2}
& = frac 14 bigg[ln bigg|frac{x-1}{x+3} bigg| bigg ]_2^infty \
& = frac 14bigg(ln(1) - lnBig(frac 15 Big) bigg) \
& = -frac 14 ln Big(frac 15Big) \
& = frac 14 ln5
end{align}
Basically, there are many subtle things going on here.
For example, I assume you found that the integral diverges because you were trying to evaluate at $(1)$. However, the reason that it ends up not diverging is because the "infinities cancel out".
This is clearly demonstrated when evaluating in $(2)$, where the upper limit requires computing
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg|$$
We know that
$$lim_{xrightarrow infty} bigg|frac{x-1}{x+3} bigg| = 1$$
Hence
$$lim_{xrightarrow infty} ln bigg|frac{x-1}{x+3} bigg| = ln(1) = 0$$
which is where the "infinities cancel out".
This tells us that we should be very careful when using the rules of integration. For example, we very commonly use the fact that
$$int_a^b big(f(x)+g(x) big) dx = int_a^b f(x)dx + int_a^b g(x)dx$$
However, this is true if and only if both of the integrals on the right hand side converge. In this case, that is not true, thus we cannot split the integrals.
answered Nov 27 at 0:50
glowstonetrees
2,195317
edited Nov 27 at 0:56
answered Nov 27 at 0:50
glowstonetrees
2,195317
answered Nov 27 at 0:50
glowstonetrees
2,195317
answered Nov 27 at 0:50
glowstonetrees
2,195317
2,195317
add a comment |
add a comment |
up vote
0
down vote
When you use partial fractions you may loose convergence on each single fraction.
To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.
answered Nov 27 at 0:55
G Cab
17.4k31237
add a comment |
up vote
0
down vote
When you use partial fractions you may loose convergence on each single fraction.
To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.
answered Nov 27 at 0:55
G Cab
17.4k31237
add a comment |
up vote
0
down vote
up vote
0
down vote
When you use partial fractions you may loose convergence on each single fraction.
To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.
answered Nov 27 at 0:55
G Cab
17.4k31237
When you use partial fractions you may loose convergence on each single fraction.
To bypass that, put the upper integration limit as $n$, take the partial fractions, integrate them, recompose the sum, take the limit $n to infty$.
answered Nov 27 at 0:55
G Cab
17.4k31237
answered Nov 27 at 0:55
G Cab
17.4k31237
answered Nov 27 at 0:55
G Cab
17.4k31237
answered Nov 27 at 0:55
G Cab
17.4k31237
17.4k31237
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3015158%2fimproper-integrals-with-polynomial-in-the-denominator-how-does-p-series-apply%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
